A Brief Introduction to Linear Response Theory with Examples in Electromagnetic Response

Transcription

1 A Brief Introduction to Linear Response Theory with Examples in Electromagnetic Response Robert Van Wesep May 3, 2013 In order to gain information about any physical system, it is necessary to probe the system. Furthermore, we should require that any probe be sufficiently weak so that we can think of the system as independent of the probe. Of course, this is an idealized situation. It is always the case that in order to treat an experiment exactly the probe and the system must be considered as a whole. However, for a vast number of experiments the approximation that a weak probe perturbs a system in such a way that complex interactions between the probe and the system can be ignored turns out to be more than adequate. The connection of the theory of a system to the experiments performed on that system is the object of linear response theory. As such, it provides a vital and powerful tool for the communication between the experimental and theoretical communities. I will begin by framing the problem in a very general way, following Fetter and Walecka 1, in terms of the state vector of the system of interest in the Schrödinger picture. Such a state satisfies the Schrödinger equation (in units such that = 1): i Ψ(t) = Ĥ Ψ(t) (1) t Now, consider an external, time-dependent probe characterized by the Hamiltonian Ĥext(t) that is turned on at a time. The modified state vector 1 A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, pp , Dover Publications, Inc., Mineola, New York,

2 satisfies the new Schrödinger equation: i t Ψ(t) = ( Ĥ + Ĥext(t) ) Ψ(t) (2) This equation is very complex and in general difficult to solve. Even if we can solve (1), solving (2) will likely be impossible. Luckily, we are not trying to solve this problem exactly. As I described above, we are interested in small external probes. Thus, we may try to use what we know, namely the time evolution operator of the unperturbed system: Hence, I will use the ansatz with the boundary condition Û(t) = e iĥt Ψ(t) = Û(t)Â(t) Ψ(0) (3) Â(t) = 1 for t < and see if I can find a solution that is correct in the limit of small perturbations. Plugging (3) into (2) ( iû(t) i tû(t)â(t) Ψ(0) = ( ) Ĥ + Ĥext(t) Û(t)Â(t) Ψ(0)  ) iĥâ(t) + Ψ(0) = ( ) Ĥ + Ĥext(t) Û(t)Â(t) Ψ(0) t iû(t)  Ψ(0) = Ĥext(t)Û(t)Â(t) Ψ(0) t we can deduce a differential equation for Â(t): i  t = Û (t)ĥext(t)û(t)â(t) (4) In this expression we can recognize in this expression the external Hamiltonian in the Heisenberg picture: Ĥext(t) H = Û (t)ĥext(t)û(t) and so I will write (4) compactly as i  = ĤH t ext(t)â(t) (5) 2

3 In order to tackle the solution to this equation we first convert it into an integral equation: t i dt  t = dt Ĥ t ext(t H )Â(t ) Â(t) Â() = i t Â(t) = 1 i dt Ĥ H ext(t )Â(t ) t dt Ĥ H ext(t )Â(t ) This equation can be solved via iteration. Starting with Â(0) (t) = 1, it is apparent that t  (1) (t) = 1 i dt Ĥext(t H ) Iterating again we find that  (2) (t) = 1 i t dt Ĥ H ext(t ) + ( i) 2 t dt t dt Ĥ H ext(t )ĤH ext(t ) (6) But ĤH ext(t) is in some sense small, so the third term in (6) can be neglected in the limit we are considering. Thus, we have found the approximate solution for Â(t) that we were seeking. Namely, Â(t) 1 i t dt Ĥ H ext(t ) Using this result in (3) we can find an expression for the state vector of the perturbed system: t Ψ(t) ] 1 Û(t) i dt Ĥext(t H ) Ψ(0) Now that we have the state of the system we can perform a measurement. Consider an operator Ô(t) in the Schrödinger picture. Then we can look at its expectation value in the presence of the external perturbation. Ô(t) ext Ψ (t) Ô(t) Ψ(t) t = Ψ (0) 1 + i dt Ĥext(t H ) ]Û (t)ô(t)û(t) 1 i = Ô(t) + i Ψ (0) t dt Ĥ H ext(t ), ÔH (t) ] Ψ(0) 3 t ] dt Ĥext(t H ) Ψ(0)

4 Since at time t = 0 we have that Ψ(0) = Ψ H, we see that the second term is an expectation value in the Heisenberg picture. A useful special case is when Ψ H = Ψ H = Ω, the ground state of the system. Then we have δo Ô(t) ext Ô(t) = i t dt Ĥ H ext(t ), ÔH (t) ] (7) This is the response of an observable of the system to an external perturbation and now it is obvious why we used the term linear response. The response is linear in the sense that (7) is linear in the perturbing Hamiltonian. Often, the external perturbation can be written as some kind of generalized force coupled to an observable of the system. A concrete example of this that I will use for the remainder of this treatment is the coupling of an external electromagnetic field to the charge and current density of a system. This situation is more than a pedagogical exercise as it relates to a vast number of experiments including X-ray scattering, optical conductivity and Hall conductivity measurements. In this case the observable that is responding is the four-current density and the electromagnetic four-vector potential couples to the four-current density. Thus I will write H ext as H ext (t) = d 3 x ĵ µ (x, t)a µ (x, t) Hence (7) becomes δj µ (x, t) = i t dt This can be written as δj µ (x) = d 3 x ĵ ν (x, t ), ĵ µ (x, t) ] A ν (x, t ) (8) t <t d 4 x K µν (x, x )A ν (x ) (9) where I have written t < t to remind the reader that the response of the system must occur after the perturbation. In other words, the response kernel, K µν, is retarded. It is useful to approach (9) from the perspective of path integrals as it will demonstrate that finding a general expression for K µν can be done 4

5 systematically using knowledge of the action. In the manner of Altland and Simons 2, I will define the partition function as ZA] = D( ψ, SA, ψ,ψ] ψ) e (10) If j µ couples to the four-vector potential linearly, the following relation holds j µ (x) j µ (x) = δ ln ZA] δa µ (x) A µ =0 = 1 D( Z0] ψ, ψ) δs ψ,ψ] e S δa µ = 1 D( Z ψ, S ψ,ψ] ψ) j µ (x)e This conforms to the definition of an expectation value when using path integrals. I will thus use the first line as a definition for my current density. Since we are interested in small perturbations we can Taylor expand the free energy functional: ln ZA] ln Z0] + Combining these results we find j µ (x) = d 4 x δ ln ZA] δa ν (x ) d 4 x δ 2 ln ZA] δa µ (x)δa ν (x ) A ν =0 A µ =A ν =0 A ν (x ) A ν (x ) Using (9) and assuming a neutral system with no steady-state current, we can immediately make the identification K µν (x, x δ 2 ) = ln ZA] (11) δa µ (x)δa ν (x ) A µ =A ν =0 Before continuing it is prudent to discuss how gauge invariance of the electromagnetic potential imposes constraints on the response kernel. 2 A. Altland and B. Simons, Condensed Matter Field Theory, 2nd ed., pp , , Cambridge University Press, New York,

6 Namely, the physical current j µ must be invariant under a gauge transformation of A ν. Imposing this on (9) we find that d 4 x K µν (x, x )A ν (x ) = d 4 x K µν (x, x ) ( A ν (x ) + x ν f(x ) ) t <t t <t = d 4 x K µν (x, x ) x ν f(x ) = 0 t <t = ( d 4 x x ν Kµν (x, x ) ) f(x ) = 0 t <t = ν x K µν(x, x ) = 0 (12) In addition, conservation of current must be imposed on (9): xj µ µ (x) = 0 = d 4 x xk µ µν (x, x )A ν (x ) = 0 t <t = µ xk µν (x, x ) = 0 (13) Note that we also could have deduced (13) from (12) by noting that (11) = K µν (x, x ) = K νµ (x, x). Now we can turn to deriving the electromagnetic response kernel using (9). The beauty of this formulation is that the functional derivatives in (9) are only with respect to the external potential. So, we can ignore all of the complex interaction terms in the action and focus only on the parts that contain the potential. I will ignore interaction of spin with the electromagnetic field and thus write the relevant action as S em = d 4 x ψ(x) eφ(x) + 1 ( ) 2 ] p ea(x) ψ(x) 2m = = = d 4 x ψ eφ e ] 2m (p A + A p) + e2 A 2 ψ 2m d 4 x ψ eφ + ie ] 2m ( A + A ) + e2 A 2 ψ 2m d 4 x eφ ψψ + ie ( ψ 2m ψ A + A ψ ψ ) ] + e2 A 2 2m ψψ where in the second line I dropped a term that did not depend on the potential. Now, I will place the scalar and vector potentials back into a 6

7 four-vector A µ = (φ, A) and identify the four current as j µ = (j 0, j) where j 0 = e ψψ ) and j = 2m( e i( ψ ψ ψ ψ) ea ψψ. Hence, I will write Sem in a convenient form: S em = d 4 x A µ j µ (14) It is important to remember that in (14) j µ depends on A µ. Now we can write (10) as ZA] = D( ψ, ψ) e {S 0 ψ,ψ]+s ema, ψ,ψ]} and start to take the functional derivatives in (11) 3 : δ δa ν (x ) ln ZA] = 1 D( Z ψ, δs em ψ,ψ] ψ) δa ν (x e SA, ) = 1 D( Z ψ, ψ) d 4 x A µ (x) δj ] µ(x) δa ν (x ) + j ν(x SA, ψ,ψ] ) e δ 2 δa µ (x)δa ν (x ) ln ZA] = 1 D( Z ψ, ψ) 2 δj ] µ(x) δa ν (x ) + j ν(x SA, ψ,ψ] )j µ (x) e 1 δz D( Z 2 δa µ (x) ψ, ψ) d 4 x A µ (x) δj ] µ(x) δa ν (x ) + j ν(x ) e SA, ψ,ψ] Where I have used the symmetry of simultaneously exchanging µ & ν and x & x in the last equality. Now we can evaluate this for A µ = A ν = 0, which obviously eliminates the 1 Z 2 δz factor. δa µ (x) δ 2 ln ZA] δa µ (x)δa ν (x ) = 1 Z = A µ =A ν =0 D( ψ, ψ) e S 0 ψ,ψ] 2 δj µ(x) )] δa ν (x ) + j µ(x)j ν (x 2 δj µ(x) δa ν (x ) + j µ(x)j ν (x )] A µ =A ν =0 A µ =A ν =0 3 Note that Altland and Simons make several mistakes in their derivation that combine to give the correct result. The derviation presented here is, to the best of my knowledge, correct. 7

8 We must be careful about the δjν term because the current depends on δa µ the potential. I will consider it in pieces. First, δjν = δj 0 = 0 for obvious δa 0 δa µ reasons. Next the non-trivial piece: δj k (x ) δa i (x) = e2 2m δ ikδ(x x ) ψ(x)ψ(x) Thus, using (11) we arrive at the final result: K µν (x, x ) = e2 2m δ µνδ(x x )(1 δ 0µ ) ˆn(x) ĵ p µ(x)ĵ p ν(x ) (15) where ˆn is the particle density operator and ĵ p µ ĵ µ A=0 is the so-called paramagnetic current density. The first term is called the diamagnetic term and the second term is called the paramagnetic term. I have written the final result as expectation values of operators, which I can do because of the connection between path integrals and propagators. This is to emphasize the fact that actually calculating these path integrals is mathematically difficult and it is often easier to think instead about operators. Of course there are some complications. First, by using partition functions I have been doing everything on the imaginary time axis. A Wick rotation must be performed to return to real time and view the thermal averages as ground state expectation values. This requires care when performing the analytic continuation. Also, the response kernel is retarded (for very good reasons!), but in order to use Wick s theorem to perform diagrammatic perturbation theory we require time-ordered products of operators. Rather than going into great detail on these issues, I will present some results for one quite important component of the response kernel. The component I will consider is K 00, called the density-density response for obvious reasons. The thermal average given by (15) is K 00 (x, x ) = ˆn(x)ˆn(x ) Motivated by (8) I will define a retarded propagator, now on the real time axis, as χ(x, t; x, t ) = iθ(t t ) ˆn(x, t), ˆn(x, t )] where the quantity in angled brackets is either a commutator or anti-commutator when considering bosons or fermions, respectively. This result can also be 8

9 obtained by considering the analytic structure of correlation functions such as those found in (15) and performing the analytic continuation rigorously. As mentioned above, we would like to be able to relate this retarded correlation function to the time ordered correlation function that we can calculate using Wick s theorem and Feynman diagrams. This can be achieved via the Lehmann representation. Rather than going into details I will simply say that the real parts of the two correlators are equal and the imaginary parts are related by Iχ(x, x ; ω) = sgn ωiχ T (x, x ; ω). χ T is the time ordered version of χ and both correlators have been Fourier transformed to the frequency domain. Let us turn our attention now to finding an expression for χ T using diagrams. In what follows I will drop the subscript T to simplify notation. We can begin by formulating an expression for χ in terms density operators in the interaction picture. χ(x, x ) = Ω T ˆn H (x)ˆn H (x )] Ω = 0 T ˆn { I (x)ˆn I (x )exp i d 4 y L int (y)} ] 0 Writing L int as a general two-particle interaction, L int (x) = 1 d 4 x V (x, x )ψ (x)ψ (x )ψ(x )ψ(x) 2 we can proceed to write an expansion for χ: χ(x, x ) = χ 0 (x, x ) + χ 1 (x, x ) + χ 2 (x, x ) + where the first couple terms are given in terms of field operators (dropping the subscript I from here on) as χ 0 (x, x ) = 0 T ψ (x)ψ(x)ψ (x )ψ(x ) ] 0 χ 1 (x, x ) = i 2 d 4 x 1 d 4 x 2 V (x 1, x 2 ) 0 T ψ (x)ψ(x)ψ (x )ψ(x )ψ (x 1 )ψ (x 2 )ψ(x 2 )ψ(x 1 ) ] 0 Now we can use Wick s theorem to proceed. Since the expectation value is between vacuum states, the normal ordered products all vanish 9

10 and only contractions remain. Starting with χ 0, the only connected, nonvanishing contraction is χ 0 (x, x ) = ψ (x)ψ(x)ψ (x )ψ(x ) = G 0 (x, x )G 0 (x, x) where G 0 (x, x ) is the propagator for a noninteracting particle going from x to x. χ 1 has a few more relevant terms given by χ 1 (x, x ) = i 2 { d 4 x 1 d 4 x 2 V (x 1, x 2 ) ψ (x)ψ(x)ψ (x )ψ(x )ψ (x 1 )ψ (x 2 )ψ(x 2 )ψ(x 1 ) +ψ (x)ψ(x)ψ (x )ψ(x )ψ (x 1 )ψ (x 2 )ψ(x 2 )ψ(x 1 ) +ψ (x)ψ(x)ψ (x )ψ(x )ψ (x 1 )ψ (x 2 )ψ(x 2 )ψ(x 1 ) } +ψ (x)ψ(x)ψ (x )ψ(x )ψ (x 1 )ψ (x 2 )ψ(x 2 )ψ(x 1 ) Where the first three terms are one-interaction irreducible and the last term is reducible. I will break the expansion of χ into irreducible and reducible parts which is schematically given by (i.e. suppressing integrals): χ 0 = χ 0 χ 1 = χ 1 + χ 0V χ 0 χ 2 = χ 2 + χ 1V χ 1 + χ 0V χ 0V χ 0 and so on. I hope the pattern is obvious and it is clear that the full expression for χ is given by the Dyson series χ = χ + χ V χ + χ V χ V χ + (16) where χ = i=0 χ i 10

11 Figure 1: Grouping the diagrams in the perturbation series by reducibility. This process is also shown diagrammatically in Figure 1. The series (16) is simple to sum χ = χ 1 + V χ + (V χ ) 2 + ] = χ 1 V χ ] 1 (17) However, the expression in (17) represents a very formal solution for χ. First of all, the inverse is non-trivial since 1 V χ depends on x and x. Also, χ includes all orders in perturbation theory. Unfortunately the interaction V often cannot be treated as small and in this case all orders must be included, an impossible task. Even so, the situation is not hopeless. There are computational methods that allow the approximation of χ, including time-dependent density functional theory (TD-DFT). In this method χ 0 is calculated using selfconsistantly obtained eigenfunctions of an effective one-particle Hamiltonian obtained using static density functional theory and V is replaced with an effective interaction. (17) then reads χ = χ 0 1 Veff χ 0 ] 1 The inversion can be performed as a matrix inversion in a computer and a numerical solution is achieved. Though it can be proved that an exact 11

12 solution is possible, in practice both the one-particle Hamiltonian and the effective interaction are approximate. But for a large class of systems TD- DFT is a powerful tool for calculating observables in solid-state systems. To go into detail about TD-DFT would take me beyond the scope of this paper, but I mention it in order to convey that theoretical calculations of condensed matter systems can be performed that relate directly with experiment. Though I have presented a brief introduction to a vast subject, I hope that I have convinced the reader that through linear response theory the quantities that we calculate theoretically and the experiments that we perform on real systems can be related in a straight forward fashion. The ability to do this is vital to connecting theory and experiment and is thus an important subject for any physicist. 12